Nonagon

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Sub Topics Regular Nonagon Irregular Nonagon Convex Nonagon Concave Nonagon Area of a Nonagon Interior Angles of a Nonagon Exterior Angles of a Nonagon Nonagon Solved Examples

A polygon with 9 sides and 9 internal angles are called as Nonagon. Nonagons are also known as Enneagon.

When all sides of the Nonagon are of equal length angles with it is called as Regular Nonagon. Normally, regular nonagon has an internal angle of 140^o.
In the regular nonagon, the measure of all the angles is also equal.
When sides if the given Nonagon are of unequal lengths, it is irregular nonagon.
The interior angles are also unequal.

The nonagon whose all interior angles measure less than 180^o is called as convex nonagon.
All 9 vertices of this nonagon will point outwards.
If one or more than one interior angle of the nonagon is a reflex angle (angle greater than 180^o), it is known as Concave nonagon.
In concave nonagon one or more vertices may point inwards.


In the above nonagon, two vertices are pointing inwards, so there are two reflex interior angles.
We can calculate area of the regular nonagon by applying three different methods.

1)    If the length of the side of the nonagon is known.

Consider one of the triangles .Draw the altitude (perpendicular ) from the apex  to the base of the triangle.(The side of the nonagon forms the base.) Let the height of the altitude be h, base of the triangle be s .


       
The angle at the apex  = $\frac{360^o}{n}$, where n: side of the nonagon  = 9

Angle at apex =$\frac{360^o}{9}$ =400       Since t = $\frac{1}{2}$ the angle at vertex t=200.

Consider tan ratio;

tan t = $\frac{(s/2)}{h}$

Solving for h,

tan t = $\frac{(s/2)}{h}$ = $\frac{s}{2h}$

h= $\frac{s}{2}$ tan t

Area of the triangle = $\frac{1}{2}$ $\times base \times height$
            = $\frac{1}{2} \times s \times h$

Consider h in terms of s,

Area = $\frac{1}{2}$$\times s \times$ ($\frac{s}{2tan t}$)

 = $\frac{s^2}{(4 tan t)}$

Since for the regular nonagon , t =20^o, tan t =tan 20^o = 0.364

So, area =  $\frac{s^2}{4}$ $\times$ 0.364  = 0.6869 $\times$ s²

As there are 9 such triangles in a nonagon, the total area of the nonagon  = 9 $\times$ area of the triangle.

Area of nonagon = 9 $\times$ 0.6869 $\times$ s²

            =6.1818 $\times$ s².


2)   When Radius r is known:

The radius of the regular polygon is the distance between the center and the vertex of the polygon (radius of the circumcircle). In the triangle, the side of the isosceles triangle is the radius. Draw the perpendicular h (height of the triangle.) Let x be the half of the base of the triangle.

Let us write x and h in terms of r.

sin t = $\frac{x}{r}$

So , x = r sin t

cos t = $\frac{h}{r}$

h = r cos t

Area of the triangle = $\frac{1}{2}$$\times base \times height$
        =x $\times$ h      (since x = $\frac{1}{2}$ base)

In terms of r,

A = r sin t $\times$ r cos t

= $\frac{1}{2}$ r² sin(2t)

{Use double angle formula for sin : sin (2t) = 2 sin t $\times$ cos t}

Total area for the nonagon  = 9 $\times$ area of the triangle

        = 9 $\times$ r²  $\times$ $\frac{sin (2t)}{2}$

Plugging t = $\frac{1800}{n}$ = $\frac{1800}{9}$ = 200

2t =400

sin 40 = 0.6428

Area =  9 $\times$ $\frac{r^2}{2}$ $\times$ 0.6428

A =    2.8925 r²


3)    When the apothem (radius of the incircle) is given

Apothem of the nonagon is the perpendicular distance from the center to the side.
 
Let a be the length of the apothem, s be the side.

Area of the triangle = $\frac{!}{2}$ $\times$ a $\times$ s

Area of the triangle = $\frac{1}{2}$$\times a \times s$

Let us show a in terms of s

tan t = $\frac{(s/2)}{a}$

so s = 2 $\times$ a $\times$ tan t

since for a regular nonagon t = 20^o,

so s=2a tan 20

    = 2 $\times$ 0.364 a

A= $\frac{1}{2}$ s $\times$ a

Plugging in s  = 0.728 a

A= $\frac{1}{2}$  $\times$ (0.728 a) $\times$ a

A=0.364 a²

So total area of the nonagon = 9 $\times$ area of the triangle

=  9 $\times$ 0.364 (a²)

=3.2757 a²
Sum of all interior angles of a nonagon can be calculated by using the formula : (n-2) $\times$ 180^oWhere, n is the number of sides of the polygon.

For the nonagon, the sum of angles = (9-2) $\times$ 180^o

Sum = 7 $\times$ 180^o

Hence, sum of interior angles =1260^o

The formula for interior angle of a regular polygon is 

: (n-2) $\times$ $\frac{180}{n}$, where n is the number of sides.

For the nonagon plug in n = 9

Interior angle = (9-2) $\times$ $\frac{180^o}{9}$

 =7 $\times$$\frac{180^o}{9}$ = 140^o
For exterior angle, use the formula $\frac{360^o}{n}$, where n is the number of sides of the polygon.
Plug in n=9, exterior angle = $\frac{360^o}{9}$ = 40^o